Gauge Invariance in Quantum Mechanics for Charged Particle

Question

In Sakurai's book chapter 2, he has discussed the diffence between canonical momentum and kenitic momentum under gauge transformatiom. The former should be gauge-dependent, and the latter one would be gauge invariance. What confuses me is that when he try to construct operator that relates the transformed state $|\tilde{\alpha}\rangle$ to original state $|\alpha\rangle$, he requires two things, that the epectation of position and kenitic momentum should be preserved, that is, $$\langle\alpha |x|\alpha\rangle = \langle\tilde{\alpha}|x|\tilde{\alpha}\rangle\tag{2.365a}$$ and $$\langle\alpha |\left(\mathbf{p}-\frac{e\mathbf{A}}{c}\right)|\alpha\rangle = \langle\tilde{\alpha}|\left(\mathbf{p}-\frac{e\tilde{\mathbf{A}}}{c}\right)|\tilde{\alpha}\rangle\tag{2.365b}$$ Under gauge transformation $$\mathbf{A} \rightarrow \tilde{A} = \mathbf{A}+\nabla\Lambda$$ In the R.H.S of (2.365b), I don't know why the canonical momentum isn't $\tilde{\mathbf{p}}$ but still $\mathbf{p}$, that is, why here the canonical momentum seems to be same?

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My thought, but I don't know if it is correct. It seems that we can also denote the canonical momentum as $\tilde{p}$, and then compute $\langle\tilde{\alpha}|\tilde{\mathbf{p}}|\tilde{\alpha}\rangle = \langle\alpha|\mathbf{p}|\alpha\rangle+\frac{e\nabla\Lambda}{c}$

Answer 1

Consider this perspective: the respective classical hamiltonians are

$H_1(q, p) = \frac{1}{2m}(p-A(q))^2, H_2(q, p) = \frac{1}{2m}(p-\tilde{A}(q))^2 $

After canonical quantisation your Hamiltonian will be

$\hat{H}_1 = \frac{1}{2m}(\hat{p}- A(\hat{q}))^2, \; \hat{H}_2 = \frac{1}{2m}(\hat{p}-\tilde{A}(\hat{q}))^2 $

both acting on the identical underlying Hilbert space $\mathcal{H}$, but providing different time evolutions. Now one searches for an operator $\mathscr{G}: \mathcal{H} \to \mathcal{H}$ such that with $|\tilde{\alpha} \rangle := \mathscr{G}|\alpha \rangle$ it holds:

$\forall |\alpha \rangle : \hat{H}_1 |\alpha \rangle = i \frac{d}{dt} |\alpha \rangle \Rightarrow \hat{H}_2 |\tilde{\alpha} \rangle = i \frac{d}{dt} |\tilde{\alpha} \rangle $

So in that sense, the gauge transformation only affects the shape of the potential $A(\hat{q}) \neq \tilde{A}(\hat{q}) $, but not the operators $\hat{q}, \hat{p}$ themselves (in the Schrödinger Picture here).